Select an item from a set in Python

Question

I am looking to select one item from a set. It does not matter which item it is, but I want it to be the same item every time the function is called.

For example, if I had the set:

my_set = set(["ABC","ABC inc", "ABC ltd"])

Then I want to return any one of those, but the same one whenever the function is run. [ie so I would always get "ABC", or always get "ABC inc" whenever I ran with the above set.] If I was using lists, I would just take the first item, but sets don't have a first per se.

This is in contrast to selecting randomly [eg How do I pick 2 random items from a Python set? ] where it will change every time it is run.

Answer 1

What about converting to list and sorting?

my_list = list(my_set)
my_list.sort()
chosen_element = my_list[0]

Answer 2

you could use a function with memoization

def get_random(my_set,memo={}):
    if id(my_set) not in memo:
       memo[id(my_set)] = random.choice(list(my_set))
    return memo[id(my_set)]

a_set = set([1,2,3,4,5])
print get_random(a_set)
print get_random(a_set)

this would always give you the same value as long as you passed in a_set ... (a different set would give a different answer)

if you wanted to make sure the item was still in the set you could change the memo if check

def get_random(my_set,memo={}):
    if id(my_set) not in memo or memo[id(my_set)] not in my_set:
       memo[id(my_set)] = random.choice(list(my_set))
    return memo[id(my_set)]

Answer 3

I may have misunderstood your question, but a check for membership in the set should work, if you are looking for anything specific.

letter_set = set(['abc','ABC','xyz','XYZ'])
check_string = 'ABC'
if check_string in letter_set:
    output = check_string

This should only give an output value if the desired string is in the set.

Answer 4

Use min() .

>>> my_set = set(["ABC","ABC inc", "ABC ltd"])
>>> min(my_set)
'ABC'

Raises ValueError if empty. From Python 3.4 a default value can be returned instead.

Answer 5

我为此找到的另一个答案是使用 min：

elem = min(mySet)

Answer 6

Maybe this can help you:

from iteration_utilities import first
from simple_benchmark import benchmark

def forLoop(s):
    value = len(s)/2
    found = None
    for elem in s:
        if elem == value:
            found = elem
            break
    return found


def getAndSet(s: set):
    value = len(s)/2
    s.remove(value)
    s.add(value)


argumente = {2 ** i: set(range(2 ** i)) for i in range(10, 13)}
a = benchmark([forLoop, getAndSet],
              argumente,
              argument_name='set element',
              function_aliases={first: 'First'})

print(a)

Output:

 forLoop getAndSet 1024 0.000023 4.393867e-07 2048 0.000046 5.612440e-07 4096 0.000091 5.677611e-07

Answer 7

Sets are implemented as hash tables. So the order is deterministic unless you insert/remove an element.

say x = set (['1','123','2','10'])

this set will always have an order '1', '2', '123', '10' (lexicographic I guess)

but if you are inserting or removing an element then you will have to use some sort of sorting function to make the order deterministic